5. Example of substitution desired – residual whenever ( c /2) n 3 – 100 n 0 , for example, if c 200 and n 1 . desired residual 3 3 3 3 3 ) ) 2 / (( ) 2 / ( ) 2 / ( 4 ) 2 / ( 4 ) ( cn 100n n c cn 100n n c 100n n c 100n n T n T
6.
7. A tighter upper bound? We shall prove that T ( n ) = O ( n 2 ) . Assume that T ( k ) ck 2 for k < n : for no choice of c > 0 . Lose! 2 cn ) 2 / ( 4 ) ( 2 100n cn 100n n T n T
12. Example of recursion tree n 2 Solve T ( n ) = T ( n /4) + T ( n/ 2) + n 2 : T ( n /4) T ( n /2)
13. Example of recursion tree Solve T ( n ) = T ( n /4) + T ( n/ 2) + n 2 : n 2 ( n /4) 2 ( n /2) 2 T ( n /16) T ( n /8) T ( n /8) T ( n /4)
14. Example of recursion tree ( n /16) 2 ( n /8) 2 ( n /8) 2 ( n /4) 2 ( n /4) 2 ( n /2) 2 (1) … Solve T ( n ) = T ( n /4) + T ( n/ 2) + n 2 : n 2
15. Example of recursion tree Solve T ( n ) = T ( n /4) + T ( n/ 2) + n 2 : ( n /16) 2 ( n /8) 2 ( n /8) 2 ( n /4) 2 ( n /4) 2 ( n /2) 2 (1) … n 2
16. Example of recursion tree Solve T ( n ) = T ( n /4) + T ( n/ 2) + n 2 : ( n /16) 2 ( n /8) 2 ( n /8) 2 ( n /4) 2 ( n /4) 2 ( n /2) 2 (1) … n 2
17. Example of recursion tree Solve T ( n ) = T ( n /4) + T ( n/ 2) + n 2 : ( n /16) 2 ( n /8) 2 ( n /8) 2 ( n /4) 2 ( n /4) 2 (1) … n 2 ( n /2) 2 …
18. Example of recursion tree Solve T ( n ) = T ( n /4) + T ( n/ 2) + n 2 : ( n /16) 2 ( n /8) 2 ( n /8) 2 ( n /4) 2 ( n /4) 2 (1) … … Total = = ( n 2 ) n 2 ( n /2) 2 geometric series
20. The master method The master method applies to recurrences of the form T ( n ) = a T ( n / b ) + f ( n ) , where a 1 , b > 1 , and f is asymptotically positive.
21. Idea of master theorem f ( n/b ) f ( n/b ) f ( n/b ) (1) … Recursion tree: … f ( n ) a f ( n/b 2 ) f ( n/b 2 ) f ( n/b 2 ) … a h = log b n f ( n ) a f ( n/b ) a 2 f ( n/b 2 ) … #leaves = a h = a log b n = n log b a n log b a (1)
22.
23. Idea of master theorem f ( n/b ) f ( n/b ) f ( n/b ) (1) … Recursion tree: … f ( n ) a f ( n/b 2 ) f ( n/b 2 ) f ( n/b 2 ) … a h = log b n f ( n ) a f ( n/b ) a 2 f ( n/b 2 ) … n log b a (1) C ASE 1 : The weight increases geometrically from the root to the leaves. The leaves hold a constant fraction of the total weight. ( n log b a )
24.
25. Idea of master theorem f ( n/b ) f ( n/b ) f ( n/b ) (1) … Recursion tree: … f ( n ) a f ( n/b 2 ) f ( n/b 2 ) f ( n/b 2 ) … a h = log b n f ( n ) a f ( n/b ) a 2 f ( n/b 2 ) … n log b a (1) C ASE 2 : ( k = 0 ) The weight is approximately the same on each of the log b n levels. ( n log b a lg n )
26.
27. Idea of master theorem f ( n/b ) f ( n/b ) f ( n/b ) (1) … Recursion tree: … f ( n ) a f ( n/b 2 ) f ( n/b 2 ) f ( n/b 2 ) … a h = log b n f ( n ) a f ( n/b ) a 2 f ( n/b 2 ) … n log b a (1) C ASE 3 : The weight decreases geometrically from the root to the leaves. The root holds a constant fraction of the total weight. ( f ( n ))
28. Examples Ex. T ( n ) = 4 T ( n /2) + n a = 4 , b = 2 n log b a = n 2 ; f ( n ) = n . C ASE 1 : f ( n ) = O ( n 2 – ) for = 1 . T ( n ) = ( n 2 ) . Ex. T ( n ) = 4 T ( n /2) + n 2 a = 4 , b = 2 n log b a = n 2 ; f ( n ) = n 2 . C ASE 2 : f ( n ) = ( n 2 lg 0 n ) , that is, k = 0 . T ( n ) = ( n 2 lg n ) .
29. Examples Ex. T ( n ) = 4 T ( n /2) + n 3 a = 4 , b = 2 n log b a = n 2 ; f ( n ) = n 3 . C ASE 3 : f ( n ) = ( n 2 + ) for = 1 and 4( cn /2) 3 cn 3 (reg. cond.) for c = 1/2 . T ( n ) = ( n 3 ) . Ex. T ( n ) = 4 T ( n /2) + n 2 /lg n a = 4 , b = 2 n log b a = n 2 ; f ( n ) = n 2 /lg n . Master method does not apply. In particular, for every constant > 0 , we have n (lg n ) .
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